Relations - Free MCQ Practice Test with solutions, JEE Maths

Since x is the square of y in relation R.
i.e. option D is correct.

Option D: The number of relations from A to B is 2^3*2= 2⁶

Number of relation from A to B = 2^n(A)×n(B)

n(A) = p, n(B) = q
⇒ Number of relation from A to B  = 2^pq

R = {(x, y): y is a factor of x, x, y∈ N}
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R

Option D is correct.

Let m = 3 and n = 2.

The number of ordered pairs in the Cartesian product is m × n = 3 × 2 = 6.

Each relation is a subset of the Cartesian product, so the total number of relations equals the number of subsets of a set with 6 elements, which is 2⁶.

The numerical value is 2⁶ = 64, therefore Option D is correct.

X is a natural number and x < 5, the number is 1, 2, 3, 4
y = 2x

► x(1), y = 2 × 1 = 2
► x(2), y = 2 × 2 = 4
► x(3), y = 2 × 3 = 6
► x(4), y = 2 × 4 = 8

Range = {2, 4, 6, 8}

A = {1, 2, 3, 5}
R = {(x, y): x + y >10: x, y ∈ A}
For any value of x, y ∈ A, the condition does not hold.
Therefore, R = Φ (Empty set)

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

Hint: To solve this problem you don’t need to count the number of relations by assuming it on your own. Use the formula of the number of relations from set A to set B. 

Complete step-by-step answer:
 It is given that A = {1, 2} and B = {3, 4}
The number of elements in A is n(A)=2 and that of B is n(B) = 2.
The formula for the number of relations from one set to another is:
⇒2^{(number of elements in the first set) x (number of elements of the second set)} 
Therefore, we can write it as follows.
No. of relations from set A to set B = 2^n(A)n(B) =  2^(2*2) = 16 
Hence, the number of relations from A to B is 16.

When x = 1:
y = 1 + 6 / 1
y = 1 + 6 = 7
So (1, 7) ∈ R.

Again for x = 2:
y = 2 + 6 / 2
y = 2 + 3 = 5
So (2, 5) ∈ R.

Again for x = 3:
y = 3 + 6 / 3
y = 3 + 2 = 5
So (3, 5) ∈ R.

Similarly for x = 4:

For x = 4:
y = 4 + 6 / 4
y = 4 + 1.5, but this is not a natural number (∉ N).

For x = 5:
y = 5 + 6 / 5
y = 5 + 1.2, but this is also not a natural number (∉ N).

Thus, the relation R = {(1, 7), (2, 5), (3, 5)}.

Where:
Domain of R = {1, 2, 3}
Range of R = {7, 5}